The atomic mass of potassium is 39.1. What is the mass of #6.02 * 10^23# atoms of potassium?

Answer 1

#"39.1 g"#

The first thing to note here is that the atomic mass, #m_a#, of potassium, #"K"#, is not #39.1#, it is #"39.1 u"#.
The relative atomic mass of potassium, #A_r#, which is equal to its atomic mass, #"39.1 u"#, divided by #1/12"th"# of the mass of single, unbound carbon-12 atom, i.e. #"1 u"#, is equal to
#A_(rcolor(white)(a)"K") = (39.1 color(red)(cancel(color(black)("u"))))/(1color(red)(cancel(color(black)("u")))) = 39.1#
So remember, the atomic mass of an element is not a unitless quantity. Atomic masses are expressed in unified atomic mass units, #"u"#.
Relative atomic masses are unitless because they express the mass of an element relative to #1/12"th"# of the mass of single, unbound carbon-12 atom.
Now, all you have to do here is remember that one mole of a given element is defined as #6.02 * 10^(23)# atoms of that element #-># this is known as Avogadro's number.
In your case, #6.02 * 10^(23)# atoms of potassium will be equivalent to one mole of potassium. Here is where the cool part comes in.

The atomic mass of potassium is equal to

#m_a = "39.1 u"#

By definition, the unified atomic mass unit is equal to

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 u " = " 1 g mol"^(-1))color(white)(a/a)|)))#

This means that the molar mass of potassium, which tells you the mass of one mole of the element, will be equal to

#39.1 color(red)(cancel(color(black)("u"))) * "1 g mol"^(-1)/(1color(red)(cancel(color(black)("u")))) = "39.1 g mol"^(-1)#
Therefore, you can say that one mole of potassium has a mass of #"39.1 g"#.
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Answer 2

The mass of (6.02 \times 10^{23}) atoms of potassium is (39.1 , \text{g}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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